Iphone ASIFormDataRequest POST错误-iOS 6
我有以下链接到PHP文件的代码,目标是发布一些字符串并在PHP文件中检索,所有的东西都工作正常,但我的代码有问题,请参见下面的内容: iOS应用程序代码Iphone ASIFormDataRequest POST错误-iOS 6,iphone,ios,xcode,asihttprequest,Iphone,Ios,Xcode,Asihttprequest,我有以下链接到PHP文件的代码,目标是发布一些字符串并在PHP文件中检索,所有的东西都工作正常,但我的代码有问题,请参见下面的内容: iOS应用程序代码 NSString *urlString = @"http://www.myurl.com/path/ios.st.addquestion.php?"; NSURL *uploadURL = [NSURL URLWithString:urlString]; ASIFormDataRequest *request = [AS
NSString *urlString = @"http://www.myurl.com/path/ios.st.addquestion.php?";
NSURL *uploadURL = [NSURL URLWithString:urlString];
ASIFormDataRequest *request = [ASIFormDataRequest requestWithURL:uploadURL];
[request setPostValue:askerUsername forKey:@"username"];
[request setPostValue:@"1" forKey:@"questiontype"];
[request setPostValue:title forKey:@"title"];
[request setPostValue:hasAditional forKey:@"hasaditional"];
[request setPostValue:aditional forKey:@"aditional"];
[request setPostValue:hasLocation forKey:@"haslocation"];
[request setPostValue:location forKey:@"location"];
[request setPostValue:latitude forKey:@"latitude"];
[request setPostValue:longitude forKey:@"longitude"];
[request setPostValue:category forKey:@"category"];
[request setPostValue:hasImage forKey:@"hasimage"];
[request startAsynchronous];
$questiontype = mysql_real_escape_string($_GET['questiontype']);
$username_PRO = mysql_real_escape_string($_GET['username']);
$title_PRO = mysql_real_escape_string($_GET['title']);
$hasAditional = mysql_real_escape_string($_GET['hasaditional']);
$aditional_PRO = mysql_real_escape_string($_GET['aditional']);
$hasLocation = mysql_real_escape_string($_GET['haslocation']);
$location_PRO = mysql_real_escape_string($_GET['location']);
$latitude_PRO = mysql_real_escape_string($_GET['latitude']);
$longitude_PRO = mysql_real_escape_string($_GET['longitude']);
$category = mysql_real_escape_string($_GET['category']);
echo "$username_PRO";
PHP文件
NSString *urlString = @"http://www.myurl.com/path/ios.st.addquestion.php?";
NSURL *uploadURL = [NSURL URLWithString:urlString];
ASIFormDataRequest *request = [ASIFormDataRequest requestWithURL:uploadURL];
[request setPostValue:askerUsername forKey:@"username"];
[request setPostValue:@"1" forKey:@"questiontype"];
[request setPostValue:title forKey:@"title"];
[request setPostValue:hasAditional forKey:@"hasaditional"];
[request setPostValue:aditional forKey:@"aditional"];
[request setPostValue:hasLocation forKey:@"haslocation"];
[request setPostValue:location forKey:@"location"];
[request setPostValue:latitude forKey:@"latitude"];
[request setPostValue:longitude forKey:@"longitude"];
[request setPostValue:category forKey:@"category"];
[request setPostValue:hasImage forKey:@"hasimage"];
[request startAsynchronous];
$questiontype = mysql_real_escape_string($_GET['questiontype']);
$username_PRO = mysql_real_escape_string($_GET['username']);
$title_PRO = mysql_real_escape_string($_GET['title']);
$hasAditional = mysql_real_escape_string($_GET['hasaditional']);
$aditional_PRO = mysql_real_escape_string($_GET['aditional']);
$hasLocation = mysql_real_escape_string($_GET['haslocation']);
$location_PRO = mysql_real_escape_string($_GET['location']);
$latitude_PRO = mysql_real_escape_string($_GET['latitude']);
$longitude_PRO = mysql_real_escape_string($_GET['longitude']);
$category = mysql_real_escape_string($_GET['category']);
echo "$username_PRO";
当我运行应用程序并将字符串发布到PHP文件时,没有任何事情发生,就像PHP接收到空值一样,我无法找出代码中的错误,有人知道吗?过程如下:
Application=>ASIFormDataRequest(Post值)=>PHP文件接收请求并在MYSQL数据库上插入字符串(问题就在这里)。
我已经在网上和ASI网站上搜索过了,但似乎没有任何效果。我认为代码是正确的, 但是,如果您发布到PHP并在PHP中使用$\u GET,则可能会出现问题
尝试使用$\u POST获取POST值,或者您可以转储$\u请求以查看您发送给php的内容。我认为代码是正确的, 但是,如果您发布到PHP并在PHP中使用$\u GET,则可能会出现问题
尝试使用$\u POST获取POST值,或者您可以转储$\u请求以查看您发送给php的内容。您正在使用
setPostValue
来创建一个php脚本,该脚本正在查找get
变量。将所有变量的PHPcode
更改为$username\u PRO=mysql\u real\u escape\u字符串($\u POST['username'])
,或者将值作为GET
值发送给http://url?name=value&name2=value2
..您正在使用setPostValue
来创建一个PHP脚本
,该脚本正在查找GET
变量。将所有变量的PHPcode
更改为$username\u PRO=mysql\u real\u escape\u string($\u POST['username'])
,或者使用http://url?name=value&name2=value2
…您是否为请求执行委托?如果是这样,也许方法requestFailed:
可以提供更多线索。事实上,信息太少,看不清问题所在。您是否为请求
实现了委托?如果是这样,也许方法requestFailed:
可以提供更多线索。事实上,信息太少,看不出问题所在。